设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)求qa 1 a 2 +qa 2 a 3 +…+qa n a n+1
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![设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)求qa 1 a 2 +qa 2 a 3 +…+qa n a n+1](/uploads/image/z/1963153-1-3.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%85%AC%E6%AF%94%E4%B8%BAq%EF%BC%88q%EF%BC%9E0%EF%BC%89%2Ca1%3Db1%3D1%2CS5%3D45%2CT3%3Da3-b2%EF%BC%8E%EF%BC%88%E2%85%A0%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%2C%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%88%E2%85%A1%EF%BC%89%E6%B1%82qa+1+a+2+%2Bqa+2+a+3+%2B%E2%80%A6%2Bqa+n+a+n%2B1)
设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)求qa 1 a 2 +qa 2 a 3 +…+qa n a n+1
设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅱ)求
q
a 1 a 2 +
q
a 2 a 3 +…+
q
a n a n+1
设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)求qa 1 a 2 +qa 2 a 3 +…+qa n a n+1
1.
S5=5a1+5×4d/2=5+10d=45 10d=40
d=4
an=a1+(n-1)d=1+4(n-1)=4n-3
n=1时,a1=4-3=1,同样满足.(这步判断一定要有)
数列{an}的通项公式为an=4n-3.
T3=b1+b2+b3=b1(1+q+q^2)=1×(1+q+q^2)=1+q+q^2
a3-b2=4×3-3-b1q=9-q
T3=a3-b2
1+q+q^2=9-q,整理,得
q^2+2q-8=0
(q+4)(q-2)=0
q=-4(
S5=45=5a3
a3=9
a1=1
2d=a3-a1=8
d=4
an=4n-3
T3=a3-b2
b1+b2+b3=a3-b2
1+2b2+b3=9
1+2q+q^2=9
q^2+2q-8=0
(q-2)(q+4)=0
q=2
bn=2^(n-1)
第二问不太清楚意思